Nilai \( \displaystyle \lim_{x \to -1} \frac{\sin (1-x^2) \cos (1-x^2)}{x^2-1} = \cdots \)
- 1
- -1
- 2
- -2
- 0
(UM UNDIP 2010)
Pembahasan:
\begin{aligned} \lim_{x \to -1} \frac{\sin (1-x^2) \cos (1-x^2)}{x^2-1} &= \lim_{x \to -1} \frac{\sin (1-x^2) \cos (1-x^2)}{-(1-x^2)} \\[8pt] &= \lim_{x \to -1} \frac{\sin (1-x^2)}{-(1-x^2)} \cdot \lim_{x \to -1} \cos (1-x^2) \\[8pt] &= \frac{1}{-1} \cdot \cos 0 = -1 \end{aligned}
Jawaban B.